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Evaluate each limit (if it exists). Use $L$ Hospital's rule (if appropriate).$$\lim _{x \rightarrow 1} \frac{\sin \pi x}{x-1}$$

$-\pi$

Calculus 1 / AB

Chapter 27

Differentiation of Transcendental Functions

Section 7

L'Hospital's Rule

Derivatives

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

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I were asked to evaluate the limit as X approaches one of sine of pi attacks over x minus one. Now, part of the direct substitution be pretty easy. That would be the denominator Because you're substituting one and for X and 1 -1 is zero. Now the numerator is a little bit more tricky because you're you're plugging in one and for X. So you have to think about, okay, well pi times one is pie. So you have to think about where pie is on the unit circle, Which is the ordered pair -10 and sign is the Y. Uh So it is zero. So we do have that indeterminant form 0/0 to use low petals rule. Everybody's supposed to be tells I shouldn't say everybody. But there's different spellings of lumpy tiles. Um Anyway, deal limit as X approaches one stays the same. But now you have to take the derivative of sine which is co sign and using the chain rule the inside stays the same. But then you have to multiply by the derivative of PX, which would be pie And the derivative of the bottom would just be one because the derivative of -1 is zero. Um So now when we do direct substitution, all we have to do it all we have to consider. Excuse me, is the numerator and now co sign is the X coordinate at pi. So it's pi times negative one or simply negative pi is your correct answer? Mhm.

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